# Tag Info

10

Yes, it is possible. For example Kevin Brown did here and here including this table. so for the xkcd problem the answer is $-\frac{1}{2}+\frac{4}{\pi} \approx 0.773$.

10

First, Field strength. This calculation is strictly an electric potential calculation; radiation and induction are safely ignored at 50Hz. For a 200kV transmission line 20m above ground, the max electric field at ground level is about 1.2 kV/m. This number is reduced from the naive 200kV/20m=10 kV/m calculation by two effects: 1) The ~1/r variation in ...

10

AC or DC, you only get electrocuted if current passes through your body. (Current passing through any part of your body can be dangerous, and possibly cause an electrical burn, but current passing across your heart is the one that's really dangerous.) Touching just one wire at a time gives the current nowhere much to go. You are right to think that some ...

9

Electric current, by definition, is a flow of charged particles. When someone says it is the propagation of the electric field, usually he means the following: The velocity of the electrons in the wires is very slow (few cm/s if I remember it right), but when one turn on the light he doesn't see any delay. The lamp starts lighting when the electrons start ...

9

If the power line is 20m high, and has the voltage of 1MV , then the electric field (near ground), very roughly, is on order of 1000/30 kv ~ 30 000 v/m (the numbers are very approximate and the field is complicated because it is a wire near a plate scenario, and wire diameter is unknown but not too small else the air would break down, i.e. spark over, near ...

7

First, your camera is not designed to work with batteries below a certain voltage. When it detects an excessively low battery voltage it turns itself off. That circuit stays in the "off" state until voltage is completely removed from the circuit. When you operate your camera, the current required by your camera varies according to what you do with it. So ...

7

Negative current just means it flows opposite to the chosen direction. Take your example, you have a loop with increasing area and downward pointing B-field x x x x x x x x x +-------------|----------- ^ x | x x x x | x x x | y | | ---> h ^ x | x x x x | x x x | | ...

7

Virtual ground refers to a circuit element not directly connected to ground, held at a reference voltage. This reference voltage need not be the same voltage as ground either. For example many op-amp circuits were originally designed for dual power supplies (say +12V and -12V) and could handle filtering or modification of a signal that was oscillating ...

7

"Ground" refers to a particular voltage, generally taken to be "zero", or the voltage of the earth. A "virtual ground" is a wire in a circuit whose voltage is held to be zero not because it is directly connected to the true ground, but instead because it is actively driven to that voltage typically by feedback mechanisms. Here's an example of a virtual ...

7

There's an old and clichéd but actually pretty good analogy for understanding electric circuits, and that's to think of the circuit as water plowing through pipes. If you have a pipe full of water, and you turn on the tap at one end, water immediately starts flowing out of the other end. Well not quite immediately: when you turn on the tap you raise the ...

7

The three capacitors are connected in parallel. There are only two nodes in this circuit. A series connection requires at least three. The equivalent capacitance is just the sum of the three capacitances. UPDATE: The circuit can be redrawn such that the parallel connection is manifest.

6

For any given $n$, you can work it out via the rules for series and parallel resistors, but to get a general formula, valid for all $n$, doesn't look easy to me. The best way I know of is to get a recursive relationship giving the resistance of an $n$-step ladder in terms of an $(n-1)$-step ladder. If I'm not mistaken, the $n$-step ladder can be thought of ...

6

I'll take it step by step here. First I'll write the answer for the first few cases with circuit analysis. Then I'll apply a reduction to show the pattern that the problem arrives at. N=1 $$Z = R+R=2R$$ N=2 $$Z = R+\frac{1}{\frac{1}{R}+\frac{1}{R + R}} = R \left( 1+\frac{1}{1+\frac{1}{1 + 1}} \right)=\frac{5}{3} R$$ N=3 $$Z = ... 6 From an answer by Steve Selkowitz of Lawrence Berkeley National Labs (I'm quoting it in full here because the site has changed and I had to pull this from an old mailing list archive) Fluorescent Lighting - Should I turn the lights off? There have been two very resilient energy myths that have dissuaded people from turning off fluorescent lamps. ... 6 I will do the case where the material is homogeneous and isotropic, \rho = \sigma^{-1} is a constant proportional to the identity matrix. We are interested in the steady state, where none of our variables depend on time. We have \nabla \times E = 0 from Faradays law and, \nabla\cdot J = 0 from the equation of continuity, where J is the current ... 6 I'll give the answer to this question using an unusual method that showed up in the American Mathematical Monthly's problem section perhaps in the late 1970s. This is not necessarily the easy way to solve the problem, but it works out nicely from an algebraic point of view. The way most people solve most resistance problems is to use series and parallel ... 6 Power lines do cause corona discharges (power line inspection video), which produce some UV light. If the camera is picking up some UV light that might cause a few dots on the picture. If you inspect the pictures closer you will notice that the pattern only appears sometimes and only in two of the cameras of the car (the front and the rear facing camera). ... 5 Well, surely you can compute it using matrix operations. But it won't be very natural. Let me instead provide you with a very similar solution (based on a similar matrix) that you'll hopefully find useful. It's not new at all (Kirchhoff, 1847) but I think it's not very well known. I first learned about it in this Wu's review paper of Potts model, p. 252. Let ... 5 More generally, is it true that every transfer function representing an RLC-circuit network is minimum phase? I suspect the answer is true, but I am having a hard time proving it. It's not true because you can have an RLC all-pass filter. To see a more specific example, let's analyse a lattice phase equaliser topology: Writing the node ... 5 Color forces are not like electromagnetic ones. There exist no unbound color carrying particles analogous to the electron, because the forces increase with the distance rather than decrease and collective effects appear only within nuclei through residuals of the colored forces which attract the nucleons and hold them in the nuclei. Collective effects ... 5 First thing, it depends on the type of light used. For an incandescent bulb, a bit of energy is used up while turning it on; but not much. On the other hand, tubelights have inductors (choke coils) in them. These come into play when switching the tubelight on and off only. This is because inductors oppose change in current, when there is a steady current, ... 5 Although the question is not clear, my guess is that you are confused with the flow of current and mean position of electrons. In case of DC, we have a continuous flow of charge from one point to another point in the conductor, any electron completes a cycle of circuit. In case of AC, there is no net displacement of charge and this may lead one in ... 5 I don't think these distortions are necessarily caused by the power lines. Going along some way forward on your link, under the power lines and then some more, you get to this image, which shows the distortion in front of the car over an area far from the power lines, when the car is also pretty far from them. EDIT: There's also some distortion inside ... 5 I've just sacrificed an AA manganese alkaline battery to the cause of physics. When I first shorted the battery it produced a current of about 9.5 amps, which I thought was actually pretty impressive. However over the course of 30 seconds the current dropped to around 5 amps. The battery got pretty warm, though I don't think it would have set fire to ... 5 Power consumption is about linear with frequency. The processor contains millions of complementary FETs as shown. When the input goes low the small capacitance gets charged and it will hold a small amount of energy. A same amount is lost during the charging. When the input goes high again the charge will be drained to ground and be lost. So with each ... 5 In real life, the current can't jump instantaneously because there is always some finite inductance in a circuit. However, this is just a typical idealized textbook problem where the inductance is assumed identically zero, so the current can jump instantaneously according to the assumptions of the problem. Note the current also jumps in their solution for ... 5 Ohm's law \vec\jmath=\sigma\vec{E} can be derived rigorously in the limit of small electric fields using linear response theory. This leads to Kubo's formula for the electric conductivity, which relates \sigma to the zero frequency limit of the retarded current-current correlation function.$$ ...

5

When there is no resistance, as is the case with an ideal wire, any value of current satisfies Ohm's Law: $V = I R$ since both $V=0$ and $R=0$. UPDATE: But isn't V is like what causes the current? Perhaps a mechanical analogy of the resistor will help. Consider the dashpot where the velocity of the arm is analogous to current while the force acting ...

5

A battery is basically just a chemical reaction. At the negative (cathode) end of the battery the reaction releases electrons while at the positive (anode) end of the battery the reaction consumes electrons. As long as the external circuit allows electrons to flow from the cathode to the anode the reaction goes and the battery generates power. If you break ...

Only top voted, non community-wiki answers of a minimum length are eligible